Re: [ATM] Couder Masks and Zone Testing

[Guy Brandenburg <gfbrandenburg@yahoo.com>]

Michael and Byron,

Rather than messing around with trying to make a
post-script printer behave properly, I find it to be
much simpler to use a calculator, a good ruler, a
compass, and a sharp single-edge razor blade to make a
decent Couder mask. Also, one innovation I gleaned
from Michael Mills (of NOVAC) is to have all of the
'holes' in the mask be the same area by using
diameters to draw them, rather than parallel chords.
Plus, this all makes it much easier to construct the
mask on relatively durable cardboard of some sort.

The math is not hard, really. If you have a 6 inch
diameter mirror, and you want 4 zones, and you want
them to have the same area, then simply think about
what the area of the mirror is, and divide it by 4.
The area formula for a circle is pi * r^2, or 9*pi.
(It's actually much less work if you leave everything
in terms of pi.)

So each zone will have an area of 9/4 * pi, or 2.25 *
pi. That means that the radius of the first zone must
be the square root of 9/4 or the square root of 2.25,
or 1.5 inches (1 1/2 "). 

Result: you draw a circle with a radius of 1.5 inches.

The next zone needs to have the same area, but is
built around the first zone; so we can add 2.25*pi to
itself, to get 4.5*pi for the area of both regions.
And, since the area of a circle is pi * r^2, that
means that the outer radius of the second zone is the
square root of 4.5 - or, if you like, the square root
of 2 times 1.5, which is about 2.12 inches. 

It will help a LOT if you have a ruler that is marked
in decimal fractions of an inch!

The third zone, to cut things short, will have a
radius that is 1.5 inches times the square root of 3.

And the fourth zone has a radius of 1.5 inches times
the square root of 4. But, hey - that's just 2. Thus,
the last zone will have a radius of 1.5 inches times
2, or 3 inches. But, hey - that means we came to the
end of the mirror.

In general, if you want to divide a mirror of radius R
(= D/2) into N zones of equal area, the first zone
will have an outer radius that has length R divided by
the square root of N, which quotient I will perversely
call Z . Then, for subsequent zones, you merely
multiply Z by the square root of 2, then the square
root of 3, then the square root of 4 (aka 2), the
square root of 5, the square root of 6, and so on,
until you get to the square root of N, at which the
square roots of N both cancel out and you have arrived
at the edge of your mirror.

As far as what portion of the zone to cut out, both
Texereau and the LeCleires suggest drawing chords that
are parallel to the diameter, and cutting between the
chords. The problem here is that the outer zones
become much, much smaller than the inner zones, and
thus are harder to read. A simpler and better solution
(IMO) that I stole from Mike Mills, as I mentioned, is
to draw two more diameters, separated from the
horizontal diameter by something like 45 degrees. Then
cut out alternate swaths of the zones that are
alternately above and below the horizontal diameter,
but are contained between the two other diameters.

There is no way I can draw this in ASCII art, but if
you want I could send you a bitmap image that
illustrates the general idea.

Guy Brandenburg



--- Michael Lindner <mikell@optonline.net> wrote:
> On Wednesday 14 April 2004 01:37 pm, Byron wrote:
> > I have read some of about creating a Couder mask,
> but right now I would
> > like some type of software or anything that will
> make creating the masks
> > easier. I can always grasp a greater understanding
> of it as I go along.
> 
> Tex will also generate a printable couder mask.
> 
> http://home.att.net/~mikel/
> 
> -- 
> Michael Lindner
> http://www.starastronomy.org ***
> http://home.att.net/~mikel
> http://www.atmsite.org *** http://www.atmlist.net
> 
> _______________________________________________
> ATM mailing list http://www.atmlist.net/


=====
Guy  BrandenburgWashington, DChttp://home.earthlink.net/~gfbranden/GFB_Home_Page.html


	
		
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